Relative to the origin $O$, two fixed points, $A$ and $B$, have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. The point $P$ has position vector $\mathbf{p}$ given by $\mathbf{p}=\lambda\mathbf{a}+(1-\lambda)\mathbf{b}$ where $\lambda$ is a parameter such that $0<\lambda<1$.

1. Express the displacement vector $\vec{AP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

$\vec{AP} = \mathbf{p} - \mathbf{a}$
$= [\lambda\mathbf{a} + (1-\lambda)\mathbf{b}] - \mathbf{a}$

2. Factorise the expression to show it is a scalar multiple of $\vec{AB}$ and conclude collinearity.

$\vec{AP} = (\lambda - 1)\mathbf{a} + (1-\lambda)\mathbf{b}$
$= (1-\lambda)(\mathbf{b} - \mathbf{a})$
Since $\vec{AB} = \mathbf{b} - \mathbf{a}$, we have:
$\vec{AP} = (1-\lambda)\vec{AB}$
Since $1-\lambda$ is a real scalar, $\vec{AP}$ is parallel to $\vec{AB}$. Because point $A$ is common to both vectors, the points $A$, $P$, and $B$ are collinear.

1. Set up the scalar product equation using the condition that the foot of the perpendicular $P$ lies on the line $AB$.

From part (a), any point $P$ on the line $AB$ can be expressed in the form $\mathbf{p} = \mathbf{a} + \mu(\mathbf{b} - \mathbf{a})$, where $\mu = 1 - \lambda$.
Since $OP$ is perpendicular to $AB$, their scalar product must be zero:
$\vec{OP} \cdot \vec{AB} = 0$
$(\mathbf{a} + \mu(\mathbf{b} - \mathbf{a})) \cdot (\mathbf{b} - \mathbf{a}) = 0$

2. Expand the scalar product and apply the condition that $\angle AOB = 90^\circ$.

Given $\angle AOB = 90^\circ$, vectors $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, so $\mathbf{a} \cdot \mathbf{b} = 0$.
Expanding the dot product equation:
$\mathbf{a} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{a} + \mu(\mathbf{b} \cdot \mathbf{b} - 2\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{a}) = 0$
$0 - |\mathbf{a}|^2 + \mu(|\mathbf{b}|^2 - 0 + |\mathbf{a}|^2) = 0$
$-|\mathbf{a}|^2 + \mu(|\mathbf{a}|^2 + |\mathbf{b}|^2) = 0$

3. Solve for $\mu$ and substitute it back into the position vector equation to show the required result.

$\mu(|\mathbf{a}|^2 + |\mathbf{b}|^2) = |\mathbf{a}|^2$
$\mu = \dfrac{|\mathbf{a}|^2}{|\mathbf{a}|^2 + |\mathbf{b}|^2}$
Substituting this back into our expression for $\mathbf{p}$:
$\mathbf{p} = \mathbf{a} + \dfrac{|\mathbf{a}|^2}{|\mathbf{a}|^2 + |\mathbf{b}|^2}(\mathbf{b} - \mathbf{a}) \text{ (shown)}$